Question 115
Show that for positive numbers $x, y, z, a, b, c$ so that $x + y + z = a + b + c$ the following inequality holds:
$$
\frac{x}{a+b} + \frac{y}{b+c} + \frac{z}{c+a} + \frac{a}{x+z} + \frac{b}{x+y} + \frac{c}{y+z} \gt 2.
$$
$$
\frac{x}{a+b} + \frac{y}{b+c} + \frac{z}{c+a} + \frac{a}{x+z} + \frac{b}{x+y} + \frac{c}{y+z} \gt 2.
$$